Problem: The tangent line to the graph of function $f$ at the point $(-1,5)$ passes through the point $(-7,7)$. Find $f'(-1)$. $f'(-1)=$
The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $f'(-1)$ gives the slope of the tangent line to the graph of $f$ where $x=-1$, which is the point $(-1,5)$. We know this line passes through $(-1,5)$, and we are also given that it passes through $(-7,7)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{7-5}{-7-(-1)} \\\\ &=\dfrac{2}{-6} \\\\ &=-\dfrac13 \end{aligned}$ In conclusion, $f'(-1)=-\dfrac{1}{3}$.